CUET · MATHS · PYQ PAPER 2025
Match List-I with List-II
Consider the function \(f(x)=2 x^3-21 x^2+36 x+80, x \in[0,6]\).
| List-I | List-II |
| (A) one of its critical points is at \(x=\) | (I) \(-28\) |
| (B) Its absolute maximum value is | (II) \(-42\) |
| (C) Its absolute minimum value is | (III) 97 |
| (D) Its second derivative at \(x=0\) | (IV) 6 |
- A (A) - (I), (B) – (III), (C) - (II), (D) - (IV)
- B (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
- C (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
- D (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
Answer & Solution
Correct Answer
(C) (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Step-by-step Solution
Detailed explanation
\(f'(x) = 6x^2 - 42x + 36\) \(6x^2 - 42x + 36 = 0 \implies x^2 - 7x + 6 = 0 \implies (x-1)(x-6)=0\) (A) Critical points: \(x=1, x=6\). One critical point is \(x=6\). \(f(0)=80\) \(f(1)=2(1)^3-21(1)^2+36(1)+80 = 2-21+36+80 = 97\)…
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